 Well, I would like to warmly thank Pepe, John, Bertrand and Todd for this beautiful conference. I'm trying to buy Todd's gentleness by thanking him and Well, I'm having a great time, and I'm pretty sure I'm not the only one so Okay, so I'm going to tell you about representations of the fundamental group of the whitehead link complements of WLC was will always stand for whitehead link complement in my talk and I would like first as an introduction to to motivate what I'm going to tell you After so here is a an example of Something Nice That gamma Be a subgroup of pu21 discreet, so This is already nice in itself and Well gamma, oh, sorry, I'm breaking all the jokes. I Need more of them so gamma acts on The complex hyperbolic two space H2c which we can think of as the unit ball In CP2 Equipped with a complex hyperbolic metric Well, and actually I want to put a closure on that So it acts on the closure of the unit ball, which is Obviously the unit ball and its boundary and its boundary is S3 and So as gamma is discreet, so we have a We have a limit set lambda gamma, which is contained in S3 and I want to look at the complement of this limit set when it's not empty so Omega gamma and I'm going to look at the complement in S3 So in S3 So Gamma acts Sorry, I'm I'll try to be gentle with the jokes discontinuously Omega gamma So of course omega could be empty For instance if lambda gamma is a lattice, but I'm not going to consider this case I'm going to consider the case where omega gamma is non-empty and when that's the case we can look at the quotient of Omega by the action of gamma and this is Well, this can be a three manifold or maybe it can be a three all default But it's something Like that let's say a three manifold and Well, it is naturally equipped with a certain structure Which is called a spherical CR structure So I'm not going to be very specific about about this But let's say it's the structure that has a manifold like that So CR stands for Cauchy Riemann and it's a general structure for say real sub manifold of a complex vector space for instance and Spherical is just referring to S3. So that's a manifold that has the structure of S3 as a real sub manifold of C2 and Well before going any further Maybe I'd like to give you a few simple examples. The first one is going to be when Gamma is a finite group generated by one elliptic element where But I'm not just taking any elliptic element. So let me make a drawing So this is going to be H2C and this elliptic element preserves two Orthogonal complex lines. So I'm drawing in four dimensions Well, I will do my best. So these two potatoes are Actually discs seen in some sort of perspective so contrary to the drawing the intersect in only one point Which is that point and the elliptic element acts on both these complex lines as a rotation And I'm going to assume that it acts as a rotation through an angle 2 pi over p and here 2 pi Over q so it has finite order And if you look at the quotient of S3 by the action of this group what you get is a length space So this is an example of a manifold That you obtained. Yes Sorry Well here What happens is that you have a fixed point inside the ball, but you have no fixed point on the boundary So that's an issue It can very well be the case that you have elliptic elements in the group that have fixed point inside the ball But not outside. So it's not there's no contradiction. It can be a manifold. Well, it's just a finite group. So There's yeah. Yes Yeah, I'm Okay, then a second example which I'm going to be less Specific about is when gamma is generated by a single loxatronic map and in that case you get S2 times S1 so you have to think a little on it, but that's not very complicated And a third example, which is slightly more involved is when for instance gamma Is a surface group preserving a disc An embedded disc. So for instance it can preserve such a such a complex line or a real plane or Something more complicated, but in these cases the result is Well the quotients of the ball by this is a disc bundle over the surface and if you look at the boundary you get a Circle bundle where sigma is the surface now a fourth example Which is more complicated is due to rich Schwartz. So it's around 2000 2002 maybe Schwartz came up With an example Where omega gamma over? gamma is homeomorphic to the whitehead link Compliment so at some point I'm going to draw a whitehead link complement, but I need to I Want to do this as late as possible because I want to have an excuse to do to not being able to Drawing a whitehead link complement. Okay, so this is a very interesting object because what we obtain here is Manifold a complex hyperbolic manifold Which is the quotient of the ball by the action of gamma of which boundary is the whitehead link complement Which is a real hyperbolic manifold. So we have an object which is real hyperbolic on the outside and complex hyperbolic in the inside and Schwartz give delicious name to these things Which is row? Chi so That's the the acronym of his PDF file actually if you look at it. So that's the rocky Manifold Okay, and it seems a very natural question to wonder Which hyperbolic manifold can one obtain this way? so now these spherical CR structures while they can be thought of in terms of XG structures where X is S3 and G is pu21 Yes Well, I'm being slightly Unprecise on that because having a structure's well The picture I described in the beginning is More the one of a uniformization and it could be that I mean there they can have structures So in this way, I was more thinking of a uniformization But some of them can admit structures that are not uniformization. So that's You meet gamma here Well, the quotient is a hyperbolic manifold in this it's topologically a hyperbolic manifold meaning that we could put on it Hyperbolic structure, but this structure is obvious. It's clearly not hyperbolic. It's something different. Yeah It's it's not always this way I'm not hyperbolic So that's why Schwartz example was so exciting and so we have XG structures where so X is X3 and G is pu21 and Whenever you have such an XG structures you have a holonymy representation which is representation row of Pi 1 of the manifold you're putting the structure on to the structure group G and so for that reason if you're looking for structures, you can begin by looking for representations and well This is actually what So that there's a program initiated by by Falbell which Looking which is Well, the question is to try finding representations of fundamental group of three manifold into well Maybe the initial motivation was pu21 but then looking for such presentation to PSL 3c or Pgl 3c and So the general idea of this of this program is to somehow mimic Thurston's construction of gluing of tetrahedra in this frame so you have to adapt it of course and it's a long and difficult Road and So there's There's a few people I should add so Falbell Guillieu Koselef Rouillet Bergeron, so sorry. I'm not respecting the There are multiple papers and so I'm not respecting the alphabetical order And I'm probably forgetting some a few names and Well from from this program came up a certain number of representations. Yes Yes, there are I was kind of sleeping them under the carpet, but Schwartz after Well, he came up with a closed example, but it's I mean, I don't really understand it. I'm at a must admit so yeah and and So They came up with a with a family of representations and already finding a representation of the fundamental group of a of a three manifold of hyperbolic three manifold to SL3c Is very difficult and then once you have this representation you have to study them and see What can we say about them and for instance are they discrete give give do they give structures and this sort of questions and so Maybe the first representations representation that what that was found is a representation of the figure 8-0 complement and A few years ago dough and Falbell found a Uniformization sorry Well uniformization is basically Basically that so that's that's a representation which is discrete So you have an additional condition that the elliptic fixed point Inside the space should be isolated. So this is a way to avoid certain elliptic elements That have that fix globally complex lines So we don't want them and then of course you want that the the image of the discontinuity region at infinity by the group The quotient is homeomorphic to the manifold and so Well, that's what Doro and Fabelle did for the figure 8-0 in the so the first example was the one of the white head link complement and for many years it was Not really known if the figure 8-0 had a such a uniformization and it turns out that yes Then Doro Showed that these uniformizations can be deformed. So it's a flexibility result No, they're not. I mean there's absolutely no hope for this representation to be faithful because think of it If you're in S3 the limit set is typically a curve a Circle say so the complement of the circle is not Simply connected. So if you make a I mean it cannot be faithful. In fact, there will be elliptic elements So there is no chance that you will do this by Well, no chance. Well, it's a little more complicated to Sorry, I don't want to say that it's impossible. I just want to say this is not the case here and all the examples we know are non-faithful and So about the connection between these structures and representations So this is a sorry, I should say 2013 I think that's 2014 correct me if I'm wrong and Doro proved in fact that there existed two hyperbolic three-manifolds say M and hen hyperbolic three-manifold With the property that we have two representation row one of pi one of M to pu to one row two with pi one of N To pu to one such that row one and row two are conjugate and Both give Uniformization This statement is a little weird, but okay Uniformization of M So it means if you start from the fundamental group of a surface Represent it. You show that it's discrete and then you analyze the quotient. You can have surprises it's not always the manifold you started from and Well, these two manifolds I don't want to be too explicit about them, but Well, if these names tell you something they don't tell me I mean these people are secret agents M05 am I right? 15 yeah, so I yes, but they have conjugate images No, no, no, of course. No, but I guess none of them is none of none of them is is faithful and well, so What I would like to do now is to present you a family of representations of the fundamental group of the white headling complement Which is Well, it's not simple because when you want to analyze it it turns out that it can be quite complicated But it's actually simple to describe so representations of The white headling complement so I want up so what is the white headling complements now I have to face my it's a the image of No, I mean both both so the Here's the white headling complement Well, it's almost It's almost nice and So what is the fundamental group of of this thing? so pi 1 is Generated by two elements with one relation Are of you and V and I'm going to write down this relation So are of you and V is actually the product of four commutators U V and then each time you change commutator you change one power to one element to its inverse So you get you V minus one you inverse V minus one and You inverse V? Okay, and so this element is a Is is is trivial So if you want to find a representation of the fundamental group of the white headling complement to pu to one It means you have to well or to s u to one which is a triple cover of pu to one. So it's not big deal You have to find two matrices in s u to one that satisfy this relation and a priori this is Really complicated because it's polynomial equations among the coefficients of this matrix. So it's it's difficult. So here is a an observation if you look at the free product of two copies of Z over 3 z well, it is a quotient of I'm going to you know this group pi and Well, we can quickly prove this because it's a it's a simple exercise. Well, no, it's not it's not exactly true but What I'm going to do is to write u equals to s t well s and t are the generators of this product and V equals T s t why not and If I compute this r of s t t s t is equal to the commutator of s t and s inverse t minus 3 s inverse squared well, that's a computation and clearly This is zero in the if s and t have both of the three Okay, so in fact it tells us that Okay, so sorry Okay, so clearly If I do this I define a morphism from this group here to that group here And if I mod out by the kernel of this morning more morphism, I see that This guy here is a quotient of that that guy here and the consequence. I'm interested in is the following any group generated by two elliptic Elements in pu2 1 is representation of Pi so that's nice because it gives us a lot of examples and I'm going to Say a little more about the set of these groups and maybe the first remark I would like to To do is that Schwartz's group belong To that family So it is generated by two other three elements and of course this representation is not faithful. Yes Yo, okay of Groups generated by two elliptic elements, but I'm going to say a little more about I mean I'm not exactly sure of what you mean. Yes Yes, okay, maybe we can discuss it after that, okay So Schwartz's group belong to that family Okay, so now what is a another three elliptic element in su2 1 so the typical example is this That's a 3 by 3 matrix and I want 1 omega omega squared Where omega is a cube root of of 1 so and Well, obviously this matrix Has trace equal to zero and in fact the condition that the matrix in su2 1 has trace zero Characterizes this kind of all the three elliptic elements. So now I want to compute the dimension Well make a dimension count. I should say so I'm going to be very crude so pu2 1 Times well, let's say su2 1 times su2 1 has dimension 18 and So that's because su2 1 has dimension. No 16. Sorry. No, you should say something 16 as dimension 16 because su2 1 has dimension 8 has the same dimension as SL3 are Okay, now. I'm adding the condition that the trace is zero the trace is a complex number. So each condition Removes two dimensions. So I have minus 2 plus 2 So this is trace equal zero and then I want to mod out by the diagonal action of su2 1 by conjugation So I take 8 again and The result should be 4 8 plus 4 is 12. Okay So we have a four-dimensional set of such of such groups So it would be possible to put coordinates slightly more precise coordinates by geometric invariance But I won't go into it But the nice thing is that Well, it's it's a big set of representations and it's a bit of a Toy model representations, but still not so many I mean not so many is known about about these groups and Okay, so I would like to use Traces so trace facts about Su2 1 So the first proposition I would like to state Has probably been stated by John in his in his course last week the the Goldman Delta it okay, so here is a here is a I'm going to consider a matrix in su2 1 and then there exists a polynomial real polynomial such that first Well, if the trace of a is positive then a F of the trace of a sorry trace of a is a complex numbers F of the trace of a is positive then a is locks a dromek if It is negative then a is elliptic and 3 if it is zero then It means that a has to as a repeated eigenvalue and Well a is either Parabolic or Special elliptic so special special elliptic elements are those I wanted to avoid in the beginning saying that I didn't want elliptic fixed point to to accumulate and I want a To be non identity, so this is true Okay, so well, this is the f is the equivalent of x squared minus 4 in the realm of SL2 R It is slightly more Okay, okay before going any further. I just wanted to Make a remark about about this so this is a An algebraic observation you can turn it into a combinatorial observation if you know that s and t have Order three then I'm going to consider this octahedron so and When I do not be sub something it means that P is a fixed point of the something so here P is a fixed point of the product st And here this point is the fixed point of TS and then I'm going to have P S inverse T P TS inverse P STS and P T ST so there are conjugacies around these these maps for instance ST and TS are conjugated by s and these four guys here are Contrugate because s has order three, so it's just a simple computation and now if you assume that these vertices are fixed points of This plus the condition of order three it tells you that this octahedron has at least combinatorially face Identifications that match exactly with those so this is the same octahedron only PTS is there PST is the point at infinity and the four points here are the same here, so it's a flattened version of this octahedron, so it's like I'm doing some sort of stereographic projection and So it's exactly the same pattern of identifications as here Which probably tells you nothing if you're not thirst on but if you mod out this by This identification you get the white head link complement, so that's the this picture is in Thurston's note and I mean to me. It's a completely obscure. I don't see at all Why the quotient should be the white head link complement, but somehow it gives a Well, you can prove this theorem by doing this actually No, I don't think so Well, I don't know how I don't know I don't think so it Through Z3 squared you mean No, I don't think so. I don't know I Think I think it's more likely that then these representations would form a Component an algebraic component of the representation variety, but I had I mean it's it's completely Experimental at this stage side. I don't know Okay So here we are and now Here we have a Group generated by two Isometries, so here's a proposition So I'm going to consider a B in SU 2 1 and The trace of the commutator of a and B and the trace of the commutator of Of its inverse sorry are The roots of a quadratic x squared minus Sigma x of sorry, I mean I have s is there so sigma is going to be a polynomial And but I have pi there so okay forgive me. I'm going to have Slight ambiguity in my notations so of a quadratic like this where S and B are real polynomials in Trace of a trace of B trace of a B trace of a inverse B so Actually, well, this is a bit of a It's not the the most precise statement we can we can do but it tells us that the character of variety of F2 in SU 2 1 lives on this algebraic real algebraic sub variety of C 5 actually I have to or 6 depending on how I but Okay, so this comes from a similar and much more precise result on the SL 3 a character of variety of F2 Which is due to Low-town the character variety F2 SL 3c and you have a similar picture So I would like to add one more thing which is the following if I consider Phi from SU 2 1 times SU 2 1 that maps a B to Trace of a trace of B trace of a B trace of a inverse B and The trace of one of these two commutators Then Each time I have a non degenerate pairs. It is determined up to conjugation by its image. So this map Classifies pairs up to conjugation, you know, so Phi classifies pairs up to conjugation and Of course the image of this map is contained in the algebraic variety Which is here? But we don't know the image and it's a complicated subset. So it's going to be a Semi-algebraic set, but we don't know an explicit description So it classifies as long as you know that the image the images in in the image. Yeah, so it's not a very Precise result okay, and so I would like to use this to to say a little more about Groups generated by pairs of elliptics so we know That here the first two factors are going to be zero and I'm going to call Z1 and Z2 These two so I'm going to set trace of ST so here S3 equals 1 T3 equals 1 so I'm looking at pairs of all the three elliptic elements in in SU2 1 and So I have trace of S equals trace of T equals zero and trace of ST is Z1 Trace of S inverse T is Z2 and now There is one additional condition which is here that these two Traces here, so there are traces of Inverse matrices in SU2 1 and it implies that they are complex conjugate so Well, that's just the same as in S as in U3 for instance so complex conjugate so I have One additional condition, which is that the discriminant of this equation there should be negative If I want Z1 and Z2 to be satisfactory as traces of elements in SU2 1 But our product of two or three elliptic elements in this way so I'm going to compute the discriminant of The equation, so I'm not going to compute it. I'm going to tell you its value so delta is equal to F of Z1 plus F of Z2 plus Z1 squared Z2 squared minus 27 so sorry I'm considering a Sorry, that's the other way around That's a plus and that's a minus and there's a two here So F is the trace function in the previous in the previous theorem So the obvious thing to do is to get rid of these two factors by assuming that we are in this case here So I would I would like these two product ST and S inverse T to be parabolic so the good thing about doing this assumption is that The group of Schwartz has this property ST and S inverse T are parabolic and actually if you go back to here and If you draw the same picture in the real hyperbolic realm then All these should be parabolic fixed point because they are I mean Correspond to cusp of the of the of the manifold so It's somehow a reasonable thing to do so We assume ST and S inverse T are Parabolic So What does it imply? So how does a parabolic map look like so I'm going to draw a matrix again so parabolic element will be Well conjugate to an upper triangle matrix And it has this form so it's going to be e to the i theta e to the i theta e to the minus 2i Theta so it has a repeated eigenvalue and it's not diagonalizable so we have stuff here and So this is it so now if I have two Parabolic traces so the trace It's going to be equal to t2 e to the i theta plus e to the minus 2i theta and I have two parabolic traces, so I'm going to have theta 1 and theta 2 and I want to plug this in In in the in the equation there to see what I obtain so let me draw the theta 1 theta 2 plane and So well, I am not going to tell you what what we obtain if we plug This trace into so here we have zero here. We have zero, but we have to Plug this into there, but we obtain a topological disc which is slightly more square than around disc and the inside Of this disc corresponds to the condition that delta is negative And of course the boundary is where delta is zero and the outside where it's positive and So we need all the representations into su2 1 correspond to Parameters inside this disc so raps in su2 1 correspond to Parameters delta negative, but the condition that the two commutators have Conjugate traces is also true for u3 So we should be careful about that, but in fact you can check that under this assumption You have a loxodromic trace in there. So all these groups correspond to representations in su2 1 so in fact In fact Well, the map s t goes to theta 1 theta 2 is on to the closed disc So all the points inside this disc correspond to a pair of elliptic of order 3 with the additional condition that these two guys are parabolic So now the question is where Uh Where does schwarz group is where sorry that's very bad sentence. Where is schwarz's group? So schwarz's group has one unipotent boundary element So saying that it is unipotent is just saying that it's theta here is zero So I get unipotent map and actually It happens full time here And in this picture we have four copies of schwarz's group and I'm going now to state a theorem due to parker and myself If theta 1 equals theta 2 equals zero, so if we are in the center of this picture, uh, we have uh, s and t Uniformize the whitehead link complement So we have exactly the same behavior as for schwarz schwarz's example. So the the quotient of the discontinuity region in on the boundary By the action of the group is a homeomorphic to the whitehead link complement And uh Okay So I would like to add just one thing. So beyond these two points I think nothing is known in in in this disk Only maybe one one more example. I think we have four copies of a group which is contained In a in a In an arithmetic lattice in the eisenstein pick our uh lattice So we know that there are four points here. It's it's four copies of the of the same group That are discrete But it is not known if they correspond to uniformization or a structure or anything so Well structure on something probably um And beyond this I think not much is known about about this slice of the of the Uh representation variety. So how do we prove such a theorem? Well, that's about the same idea as what The dough and file bell deal did for the the white for the figure eight knot and that many people did We construct a fundamental domain and we analyze the quotient of this fundamental domain By the action of the group So we need to construct a fundamental domain inside the space And then look at its intersection with the boundary and then analyze the behavior on the boundary. This is Not simple in general because when you want to construct a fundamental domain, well, you need to Define faces for a polyhedron and in complex hyperbolic space There are no totally geodesic Faces So you have to handle. Well, there are fair substitutes for these totally geodesic spaces, which are equidistant surfaces They are not totally geodesic, but still they are relatively well understood But it means you have a it's quite complicated to understand their intersections. It involves a A lot of computations and to Finish my talk. I just would like to give a Geometric property of the boundary groups here. So if delta Is equal to zero then the trace Equations So which is the one which was there? Sorry. I erased it has two real has a real double Root the two traces of the of the of the Commutators are real and by A result with Julien paupers this implies that The group generated by s and t has indexed to in a anti holomorphic reflection group So what I mean by this is that Just like in the in the usual poincare disc. So the the full isometry group is formed by psl2 r and All the conjugates of the complex conjugation by elements of psl2 r here We have pu21 that acts by holomorphic isometries and we have another Connected component in the isometric group which is formed by conjugates of the complex conjugation like you just If you think of the unit ball in c2, you just complex conjugate the two coordinates and the complex conjugation is a isometric involution of the ball and here We can write s This is probably too low. So I'm going to come here s equals sigma one Sigma two and t equals sigma two Sigma three Where these three people here are conjugates to the complex conjugation So there are Involutions so it gives a Nice characterization of the of the boundary in geometric terms. Okay. I think I'm going to stop here. Thank you